Quadrilaterals Test

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Quadrilaterals Test - 7

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Question 1
1 / -0

ABCD is a parallelogram, and E, F, G and H are the bisectors of ∠B, ∠C, ∠D and ∠A, respectively. If HI = 2t + 19 and IF = 7t – 6, what will be the value of FH?

A
29

B
56

C
58

D
68

Solution

The bisectors of angles of a parallelogram make a rectangle.
Diagonals of a rectangle bisect each other; HI = IF.
Thus, 2t + 19 = 7t - 6
2t + 19 = 7t - 6
19 + 6 = 7t - 2t
25 = 5t
t = 25 ÷ 5
t = 5

Therefore, HI = 2t + 19
= 2(5) + 19
= 10 + 19
= 29

Also, IF = 7t - 6
= 7(5) - 6
= 35 – 6
= 29

So, HF = HI + IF
= 29 + 29
= 58
Question 2
1 / -0

In the given figure, Q and R are the mid-points of MO and ON, respectively. The area of ΔMNO is 36 cm². If MO = 3 cm and ON = 4 cm, find the length of QR.

A
QR = 3.5 cm

B
QR = 3 cm

C
QR = 2.5 cm

D
QR = 2 cm

Solution

We know that △ MON is a right-angled triangle.
MO = 3 cm
ON = 4 cm
By Pythagoras theorem,
Hypotenuse² = (base² + height²)
MN² = (ON²+ MO²)
MN² = (4² + 3²)
MN² = 16 + 9
MN² = 25
MN =
MN = 5 cm

Since Q and R are the mid-points of MO and ON, respectively;
Using mid-point theorem, QR ∥ MN
QR = of MN
QR = of 5
QR = 2.5 cm
Question 3
1 / -0

In the figure given below, DE is parallel to BC. AE = 4 cm and DB = 4 cm. If DE divides AE and EC in ratio 2 : 3, then what is the value of AD?

A
3 cm

B
cm

C
cm

D
cm

Solution

We know that DE ∥ BC; AE = 4 cm and DB = 4 cm; and DE divides AE and EC in ratio 2 : 3.
So, EC =
Now, according to side splitter theorem,

6AD = 16
AD = 16 ÷ 6
AD = cm
Question 4
1 / -0

In the given figure, ABDE is a parallelogram and F and D are the mid-points of AC and BC, respectively. If AC = 4 cm, BC = 4 cm and AB = 6 cm, then which of the following options is true?

A
FD = 2EF

B
ΔAEF ≌ ΔFCD

C
3FD = AB

D
ΔAEF ≌ ΔFDC

Solution

In the given figure, AB = 6 cm, AC = 4 cm and BC = 4 cm.
F and D are the mid-points of AC and CB, respectively.
AF = FC = 2 cm
BD = DC = 2 cm

Using mid-point theorem, FD || AB
FD = AB
FD = 3 cm

Now, in parallelogram ABDE, AB = ED = 6 cm (As opposite sides of a parallelogram are equal)
EF = ED – FD
= 6 – 3
= 3 cm

In ΔAEF and △FCD,
EF = FD
AF = FC
AE = BD = DC

Therefore, using SSS congruency rule,
ΔAEF ≌ ΔFCD
Question 5
1 / -0

In a parallelogram, the smallest angle and the largest angle are in the ratio 4 : 11. If the largest angle is decreased by 25%, what will be the new measure of the smallest angle?

A
48°

B
71°

C
81°

D
84°

Solution

We know that in a parallelogram, opposite angles are same. So, let the smallest angle in parallelogram ABCD be ∠A and the largest be ∠D.
A.T.Q,
Let ∠A be 4x and ∠D be 11x.
We know that the adjacent angles of a parallelogram always equal 180°.
So, ∠A + ∠D = 180°
4x + 11x = 180°
15x = 180°
x = 180 ÷ 15
x = 12°
Therefore, ∠A = 4x
= 4 × 12
= 48°
∠D = 11x
= 11 × 12
= 132°
Now, ∠D is decreased by 25%.
Therefore, ∠D = 132 × 0.75 = 99°
Since ∠D is decreased by 33, ∠A will increase by 33 because the sum of the adjacent angles in a parallelogram should always be equal.
Therefore, ∠A = (48 + 33)°
= 81°
Question 6
1 / -0

PQRS is a parallelogram such that PQ = 7 cm and PQ : SP = 7 : 3. Find the length of side QR.

A
7 cm

B
5 cm

C
3 cm

D
4 cm

Solution

PQ = 7 cm
PQ : SP = 7 : 3
QR = ?

( PQ = 7 cm)

SP = 3 cm
And QR = SP = 3 cm ( Opposite sides of a parallelogram are equal.)
Question 7
1 / -0

In the following rhombus ABCD, find the value of x if the ratio of the largest to the smallest angle is 3 : 5.

A
32.5°

B
37.5°

C
42.5°

D
47.5°

Solution

Clearly, by looking at rhombus ABCD, A and C are the smaller angles and B and D are the larger angles because the opposite angles of a rhombus are equal.
Therefore, A is the smallest and D is the largest.
A.T.Q.,
Let A be 3y° and D be 5y°.
So, A + D = 180° (Consecutive angles of a rhombus are supplementary = 180°)
3y + 5y = 180°
8y = 180°
y = 180 ÷ 8
y = 22.5°
Therefore, D = 5y
= 5 × 22.5
= 112.5°
Since D = 75 + x
112.5 = 75 + x
x = 112.5 - 75
x = 37.5°
Question 8
1 / -0

In a parallelogram ABCD, AC and BD are diagonals that bisect each other at O. A line BE is drawn parallel to OC and line CE is drawn parallel to OB. If the area of ΔBOC is 18 cm² and that of parallelogram ABCD is 68 cm², then which of the following statements is false?

A
Area of ΔAOD = 18 cm²

B
Area of ΔDOC = 16 cm²

C
Area of quadrilateral DBEC = 42 cm²

D
Area of quadrilateral DBEC = 52 cm²

Solution

According to the given information, the parallelogram is shown below.

We can see that OBEC is a new parallelogram formed and its diagonal is BC. Since the area of ΔBOC is 18 cm²;
Area of ΔBCE = 18 cm² (The diagonal of a parallelogram divides it into two equal triangles)
Also, in ΔBOC and ΔAOD, AD = CB (Opposite sides of a parallelogram are equal)
AO = OB and DO = OC (The diagonals bisect each other into two equal parts)
So, by using SSS congruency rule, △BOC is congruent to ΔAOD.
Therefore, ΔAOD = 18 cm²
Now, area of parallelogram ABCD = 68 cm²
Using SSS congruency rule, ΔAOB is congruent to △DOC.
Area of parallelogram ABCD = Area of (ΔAOB + ΔDOC + ΔBOC + ΔAOD)
68 = 2(Area of ΔAOB) + 18 + 18
68 = 2(Area of ΔAOB) + 36
Area of ΔAOB = 16 cm² = Area of ΔDOC
Therefore, area of quadrilateral DBEC = Area of (ΔOBC + ΔBCE + ΔDOC)
= 18 + 18 + 16 = 52 cm²
Question 9
1 / -0

In the given figure, ABCD is a parallelogram and AC is one of its diagonals.

Which of the following relationships is correct?

A
B = DCB

B
CAB = ACB

C
DAC = CAB

D
DCA = CAB

Solution

ABCD is a parallelogram.

DCA = CAB [Alternate interior angles as AB || DC]
Question 10
1 / -0

In the given triangle ABC, E is the midpoint of BC, BC = 8 cm and M and N are the midpoints of BD and DE, respectively. Find the measure of x.

A
8 cm

B
4 cm

C
2 cm

D
1 cm

Solution

E is the midpoint of BC.
Then, BE = EC
BE = BC
BE = (8) cm = 4 cm
In ΔDBE, M and N are midpoints of BD and DE, respectively.
Then, by midpoint theorem,
MN || BE and MN = BE = x = (4) cm = 2 cm
Question 11
1 / -0

In quadrilateral GSTU, if ∠G = 90° and ∠S : ∠T : ∠U = 3 : 3 : 4, then what will be the sum of the smallest and the largest angle?

A
179°

B
189°

C
191°

D
193°

Solution

Let ∠S, ∠T and ∠U be 3x, 3x and 4x, respectively.
Now, according to the angle sum property of a quadrilateral,
∠G + ∠S + ∠T + ∠U = 360°
90° + 3x + 3x + 4x = 360°
10x = 270°
x = 27°
So,
∠G = 90°
∠S = 3x = 81°
∠T = 3x = 81°
∠U = 4x = 108°
Sum of the smallest and the largest angle = 81 + 108 = 189°
Question 12
1 / -0

If MO is the bisector of ∠M and ∠O in rhombus MNOP, what will be the value of ∠NMO?

A
25°

B
27.5°

C
32.5°

D
45°

Solution

As MNOP is a rhombus, ∠NMP and ∠MPO are supplementary (because adjacent angles of a rhombus are supplementary).
∠NMP + 115° = 180°
∠NMP = 65°
As MO is the bisector of ∠M,
∠NMO = ∠NMP
∠NMO = × 65 = 32.5°
Question 13
1 / -0

In the following figure, what will be the value of ∠MNO?

A
120°

B
115°

C
110°

D
100°

Solution

∠QPM + ∠MPO = 180° (Straight line angle)
105° + ∠MPO = 180°
∠MPO = 180° - 105° = 75°
In quadrilateral MNOP,
∠PMN + ∠MNO + ∠NOP + ∠MPO = 360°
95° + ∠MNO + 70° + 75° = 360°
∠MNO = 360° - 70° - 75° - 95° = 120°
Question 14
1 / -0

What will be the measure of H in the following figure?

A
75°

B
85°

C
95°

D
105°

Solution

Quadrilateral GIHJ is a kite.
In the given figure, J and I, and G and H are diagonally opposite to each other.
J ≠ I
Thus, G = H (One pair of diagonally opposite angles is equal in measurement in a kite)
According to the angle sum property of a quadrilateral,
G + H + I +J = 360°
H + H + 95° + 55° = 360° ( G = H)
2 H = 360° – 95° – 55°
H = 105°
Question 15
1 / -0

Find the value of 'g' if IJKL is a parallelogram, and M, N, O and P are the bisectors of the angles.

A
18

B
21

C
25

D
38

Solution

According to the question,
IJKL is a parallelogram, and M, N, O and P are the bisectors.
Thus, MNOP is a rectangle as the bisectors of angles of a parallelogram form a rectangle.
Now,
3g – 36 = g + 6 (Opposite sides of rectangle are equal)
2g = 36 + 6
2g = 42
g = 21
Question 16
1 / -0

If one of the corners of a parallelogram is joined to its opposite corner, what type of triangles will be formed?

A
Right-angled

B
Congruent

C
Equilateral

D
Obtuse

Solution

In parallelogram PUSB, PU || BS, and PS is a transversal.
In ΔPBS and ΔPUS,
∠BSP = ∠UPS (Alternate interior angles)
As PU ║ BS, ∠BPS = ∠USP (Alternate interior angles)
BS = PU (Opposite sides of a parallelogram)
PB = US (Opposite sides of a parallelogram)
PS = SP (Common)
ΔPBS ≅ ΔPUS
So, if one of the corners of a parallelogram is joined to its opposite corner, it will be a diagonal which divides the parallelogram into two congruent triangles.
Question 17
1 / -0

If, in a quadrilateral, one pair of opposite angles is equal and the longest diagonal bisects the shortest diagonal into equal parts, then what type of quadrilateral is it?

A
Parallelogram

B
Rhombus

C
Kite

D
Square

Solution

A kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. The longest diagonal bisects the shortest diagonal into two equal parts.
Question 18
1 / -0

If one angle of a quadrilateral is 126° and the rest are in ratio 6 : 4 : 3, then what will be the sum of the first and second angles if arranged in ascending order?

A
126°

B
135°

C
156°

D
180°

Solution

Let the angles be 6x, 4x and 3x.
As sum of angles of a quadrilateral is 360°;
126° + 13x = 360°
Accordingly, x = 234 ÷ 13 = 18°
The angles will be:
3x = 3 × 18 = 54°
4x = 4 × 18 = 72°
6x = 6 × 18 = 108°
Given angle = 126°
Ascending order = 54°, 72°, 108°, 126°
Sum of the first and second angles = 54 + 72 = 126°
Question 19
1 / -0

Find the sum of g and z.

A
25°

B
32°

C
35°

D
46°

Solution

According to the question,
∠IGH = ∠IJH (Opposite angles of a parallelogram are equal)
5z + 25° = 105°
5z = 80°
z = 80° ÷ 5 = 16°
Similarly, 75° = 3g - 15° (Opposite angles of a parallelogram are equal)
75° + 15° = 3g
90° = 3g
g = 90 ÷ 3 = 30°
z + g = 16° + 30° = 46°
Question 20
1 / -0

What type of shape will it be, if both the diagonals of a parallelogram are of same size and are right bisectors of each other?

A
Rectangle

B
Square

C
Kite

D
Trapezium

Solution

A parallelogram has two pairs of parallel sides. The opposite sides are of equal length and the opposite angles of are of equal measure. The diagonals of a parallelogram bisect each other.

If the diagonals of a parallelogram are of same size and bisect each other at right angles, a square is formed.
Question 21
1 / -0

Study the statements carefully and choose the correct option.

Statement I: If the diagonals of a quadrilateral are equal and bisect each other, then the quadrilateral would be either a rectangle or a square.
Statement II: If ABCD is a parallelogram, E and F are the respective mid-points of AB and CD, and BD is a diagonal, then AECF is a parallelogram.

A
Only statement II is true.

B
Only statement I is true.

C
Both statements I and II are false.

D
Both statements I and II are true.

Solution

Statement I - If the diagonals of a quadrilateral are equal and bisect each other, then the shape is a rectangle.
It is true for a square and a rectangle as well. Hence, it is true.

>Statement II: If ABCD is a parallelogram, E and F are mid-points and BD is a diagonal, then AEFC is a parallelogram.

Given: E and F are the mid-points of AB and CD, respectively.
AE = AB
And CF = CD
As ABCD is a parallelogram,
AB = CD and AB || CD

Thus,

Therefore, AE = CF and AE || FC
Now, as one pair of sides is equal and parallel, AECF is a parallelogram.

Question 22

1 / -0

Read the following statements carefully and state 'T' for true and 'F' for false.

(i) For a rhombus and a parallelogram, adjacent angles are supplementary and opposite angles are equal.
(ii) If all the angles of a parallelogram become equal, then the shape thus formed is a square.
(iii) A trapezoid is isosceles if and only if the base angles are congruent.

	(i)	(ii)	(iii)
a	T	F	T
b	T	F	F
c	F	T	T
d	T	T	T

a

b

c

d

Solution

(i) For a rhombus and a parallelogram, adjacent angles are supplementary and opposite angles are equal.
For both these shapes, the statement is true as the sum of adjacent angles for these shapes is 180° and opposite angles are equal.

(ii) If all the angles of a parallelogram become equal, then the shape thus formed is a square.
It is false because it can be a rectangle or a square and we do not have any information about the length of the sides whether they are equal or not; because for a square, all the sides must be equal.

(iii) A trapezoid is isosceles if and only if the base angles are congruent.
Let ABCD be a trapezoid and AB || CD, AD = BC and DA || CE.

Hence, ADCE thus formed is a parallelogram.
Now, DA = CE and DC = AE (Properties of parallelogram)
If BC ≅ CE, then angles opposite to them are congruent.
∠CEB ≅ ∠CBE
So, by property of parallelogram and linear pair angles,
∠DAB ≅ ∠ABC
Interior angles on the same side of the transversal are supplementary.
Hence, ∠A + ∠D = 180° and ∠B + ∠C = 180°
=> ∠A + ∠D = ∠C + ∠B
=> ∠D = ∠C

Question 23

1 / -0

Fill in the blanks:

(i) For a quadrilateral ABCD, if ∠A = 90°, ∠B = 120° and ∠C = 70°, then ∠D = ____P_____.
(ii) The diagonals of a rhombus ___Q____ each other at right angles.
(iii) A kite can become a rhombusif all ____R_____ are equal.

	P	Q	R
a	70°	meet	angles
b	80°	cut	diagonals
c	80°	bisect	sides
d	60°	meet	angles and sides

a

b

c

d

Solution

For a quadrilateral ABCD,
∠A = 90°, ∠B = 120°, ∠C = 70°
We know, sum of interior angles of a triangle = 360°
=> ∠A + ∠B + ∠C + ∠D = 360°
=> 90° + 120° + 70° + ∠D = 360°
=> 280° + ∠D = 180°
=> ∠D = 360° - 280°
=> ∠D = 80°

(ii) The diagonals of a rhombus bisect each other at right angles.

(iii) In the special case where all four sides are of same length, the kite satisfies the definition of a rhombus.

Question 24
1 / -0

In the following figure, ABCD is a parallelogram. If DAC = 2(2x - 30°) and ABC = x + 30°, find the measure of BCE.

A
72°

B
68°

C
63°

D
58°

Solution

In the given figure,

DAC = 2(2x - 30°) and ABC = x + 30°
We know that the sum of adjacent angles of a parallelogram is 180°.
⇒ DAC + ABC = 180°
⇒ 2(2x - 30°) + x + 30° = 180°
⇒ 4x - 60° + x + 30° = 180°
⇒ 5x - 30° = 180°
⇒ 5x = 210°
⇒ x = 42°
Hence, ABC = x + 30°
⇒ ABC = 42° + 30°
⇒ ABC = 72°
Now, ABC = BCE (Alternate angles)
⇒ BCE = 72°

Question 25

1 / -0

ABCD is a quadrilateral, and AC and BD are its diagonals intersecting at point O.

Based on the above information, match column I with column II:

Column I	Column II
(a) If ABCD is a parallelogram and ∠ABD = 30°, then the measure of ∠AOB is	(I) 110°
(b) If ABCD is a rhombus and ∠OAB = 50°, then the measure of ∠ABO is	(II) 90°
(c) If ABCD is a parallelogram and ∠ADC = 70°, then the measure of ∠DCB is	(III) 40°
(d) If ABCD is a rhombus and ∠ODC = 40°, then the sum of ∠DOC and ∠OCD is	(IV) 140°

	a	b	c	d
1	I	II	IV	III
2	II	III	I	IV
3	III	IV	I	II
4	IV	I	II	III

1

2

3

4 Solution

(a) Given: ∠ABD = 30° = ∠DBC (Diagonal bisect the angle of parallelogram)
=> ∠ABC = 60°
Sum of adjacent angles of a parallelogram = 180°
60° + ∠BAD = 180°
=> ∠BAD = 120°
And ∠CAB = 60° (Diagonal bisects the angle of parallelogram)
Sum of interior angles of a triangle = 180°
=> ∠ABD + ∠BAC + ∠AOB = 180°
=> 30° + 60° + ∠AOB = 180°
=> 90° + ∠AOB = 180°
=> ∠AOB = 90°

(b) ABCD is a rhombus.
=> ∠AOB = 90°
And ∠OAB = 50°
Sum of interior angles of a triangle = 180°
=> ∠ABO + ∠AOB + ∠OAB = 180°
=> ∠ABO + 90° + 50° = 180°
=> ∠ABO = 180° - 140°
=> ∠ABO = 40°

(c) ABCD is a parallelogram.
∠ADC = 70°
∠DCB + ∠ADC = 180°
=> ∠DCB + 70° = 180°
=> ∠DCB = 180° - 70°
=> ∠DCB = 110°

(d) If ABCD is a rhombus and ∠ODC = 40°,
∠ODC + ∠OCD + ∠DOC = 180°
=> 40° + ∠OCD + 90° = 180° (In a rhombus, diagonals cut at 90°)
=> ∠OCD = 180° - 130°
=> ∠OCD = 50°

Required sum = 90° + 50° = 140°

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Quadrilaterals Test - 7

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Quadrilaterals Test - 7

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SHARING IS CARING

Question 1

29

56

58

68

Solution

Question 2

QR = 3.5 cm

QR = 3 cm

QR = 2.5 cm

QR = 2 cm

Solution

Question 3

3 cm

cm

cm

cm

Solution

Question 4

FD = 2EF

ΔAEF ≌ ΔFCD

3FD = AB

ΔAEF ≌ ΔFDC

Solution

Question 5

48°

71°

81°

84°

Solution

Question 6

7 cm

5 cm

3 cm

4 cm

Solution

Question 7

32.5°

37.5°

42.5°

47.5°

Solution

Question 8

Area of ΔAOD = 18 cm2

Area of ΔDOC = 16 cm2

Area of quadrilateral DBEC = 42 cm2

Area of quadrilateral DBEC = 52 cm2

Solution

Question 9

B = DCB

CAB = ACB

DAC = CAB

DCA = CAB

Solution

Question 10

8 cm

4 cm

2 cm

1 cm

Solution

Question 11

179°

189°

191°

193°

Solution

Question 12

25°

27.5°

32.5°

45°

Solution

Area of ΔAOD = 18 cm²

Area of ΔDOC = 16 cm²

Area of quadrilateral DBEC = 42 cm²

Area of quadrilateral DBEC = 52 cm²